Robbie Hatley's Solutions To The Weekly Challenge #285
For those not familiar with "The Weekly Challenge", it is a weekly programming puzzle with two parts, cycling every Sunday. You can find it here:
The Weekly Challenge for the week of 2024-09-01 through 2024-09-07 is #285. Its tasks are as follows:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Task 285-1: "No Connection" Submitted by: Mohammad Sajid Anwar You are given a list of routes, @routes. Write a script to find the destination with no further outgoing connection. Example 1: Input: @routes = (["B","C"], ["D","B"], ["C","A"]) Output: "A" "D" -> "B" -> "C" -> "A". "B" -> "C" -> "A". "C" -> "A". "A". Example 2: Input: @routes = (["A","Z"]) Output: "Z"
This is just a matter of checking for end destinations which do not also appear in other-than-last-place in the given routes. For loops and if statements will be involved.
Robbie Hatley's Perl Solution to The Weekly Challenge 285-1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Task 285-2: "Making Change" Submitted by: David Ferrone Compute the number of ways to make change for a given value in cents by using the coins Penny, Nickel, Dime, Quarter and Half-dollar. Order of coin selection does not matter. A penny (P) is equal to 1 cent. A nickel (N) is equal to 5 cents. A dime (D) is equal to 10 cents. A quarter (Q) is equal to 25 cents. A half-dollar (H) is equal to 50 cents. Example 1: Input: $amount = 9 Ouput: 2 1: 9P 2: N + 4P Example 2: Input: $amount = 15 Ouput: 6 1: D + 5P 2: D + N 3: 3N 4: 2N + 5P 5: N + 10P 6: 15P Example 3: Input: $amount = 100 Ouput: 292
This task is a bit more complicated than the first, but not really hard. The trick is to first consider what numbers of half-dollars one can use; then for each of those numbers, what numbers of quarters one can use; then for each of those numbers, what numbers of dimes one can use; then for each of those numbers, what numbers of nickels one can use; then the number of pennies needed will be forced by those 4 numbers. Nested for loops makes this easy.
Robbie Hatley's Perl Solution to The Weekly Challenge 285-2
That's it for challenge 285; see you on challenge 286!
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